First year Algebra:
eigenvalues and eigenvectors
Let A be a n  n matrix.
 is an eigenvalue for A if there exists a vector u  0 such that
Au  u.
If such a vector u exists, it is said to be an eigenvector associated with the
eigenvalue .
(I) How to find the eigenvalues and eigenvectors
Suppose that  is an eigenvalue for A .
 x1 
 
Then there exists a non-zero vector u   ...  such that Au  u or,
x 
 n
equivalently, such that ( A  I n )u  0 , where I n is the n  n identity matrix.
Since ( A  I n )u  0 has a non-trivial solution u , the matrix ( A  I n ) is not
invertible, i.e. A  I n  0.
Note that A  I n is a polynomial in  , so you get the eigenvalues of A by
finding the roots of A  I n .
Say  0 is such an eigenvalue. In order to find the eigenvectors associated
with  0 , you have to solve the system Au  0 u for x1 ,..., xn .
(II) Example: two by two matrices
  3 2
 .
Let A  
  2 1
To find the eigenvalues of A , form the matrix A  I 2 , find its determinant
A  I 2 and solve the equation A  I 2  0 :
2 
  3 2
 1 0   3  
   
  
.
A  I 2  
1   
  2 1
 0 1   2
Now,
A  I 2  (3   )(1   )  (2)(2)  2  2  1 ,
so that the eigenvalues of A are the roots of 2  2  1  (  1) 2 , i.e. A has a
repeated eigenvalue: 0  1 .
Now we have to solve the system Au  0 u . Here 0  1 , so that
Au  0 u 
Au  u ,
which yields
x 
  3 2  x1 

    1 
  2 1  x2 
 x2 
or, equivalently,
  2 2  x1   0 

    .
  2 2  x2   0 
x 
A vector u   1  is therefore an eigenvector associated with the eigenvalue
 x2 
0  1 if and only if its coordinates satisfy
 2 x1

 2 x1
 2 x2  0
 2 x2  0
,
i.e. if and only if x1  x2 .
Hence the eigenvectors associated with the eigenvalue 0  1 are of the
1
form u     , where  is a real number.
1
(III)
Example: three by three matrices
1
  3 1 1
 3  



Let A    7 5  1  . Then A  I 3    7
5
  6 6  2
 6
6



A  I 3  (3   )
5
6
1 

 1  and
 2   
1
7
1
7 5


 3  12  16 .
2 6 2 6
6
In general, it is not easy to find the roots of a polynomial of degree 3. We try
some simple values and hope we will be lucky. For   2 , you get
A  I 3  (2) 3  12(2)  16  0 ,
so that   2 is a root of A  I 3 .
Hence (  2) is a factor of A  I 3 and A  I 3 can be written in the form
A  I 3  (  2)(a2  b  c)  a3  2 (b  2a)   (c  2b)  2c.
We now have
A  I 3  3  12  16  a3  2 (b  2a)   (c  2b)  2c,
so that a, b and c must satisfy
 a  1
 b  2a  0

,

c  2b  12
 2c  16
which yields
a  1

b2.
 c8

We are now in position to find all the factors of A  I 3 :
A  I 3  (  2)(2  2  8)  (  2) 2 (  4).
A therefore has two eigenvalues: 0  2 and 1  4 .
To get the eigenvectors associated with 0  2 , we solve ( A  2I 3 )u  0 :
  1 1  1 x1   0 
  x1

   

( A  2 I 3 )u    7 7  1 x2    0    7 x1
  6 6 0  x   0 
 6 x
1

 3   

 x2
 7 x2
 6 x2
 x3  0
 x  x2
 x3  0   1
.
x3  0

0
Hence the eigenvectors associated with the eigenvalue 0  2 are of the
1
 
form   1 .
 0
 
To get the eigenvectors associated with 1  4 , we solve ( A  4I 3 )u  0 :
  7 1  1  x1   0 
 7 x1

   

( A  2 I 3 )u    7 1  1  x2    0    7 x1
  6 6  6  x   0 
 6 x
1

 3   

 x2
 x2
 6 x2
 x3  0
 x 0
 x3  0   1
.
x 2  x3

 6 x3  0
Hence the eigenvectors associated with the eigenvalue 1  4 are of the form
 0
 
  1 .
 1
 
(III)
Exercises
Find the eigenvalues and the eigenvectors of the matrix A , where
(i)
 0 2

A  
 3 5
(ii)
 0 5

A  
 0 1
(iii)
 1  3 3


A   3  5 3
 6  6 4

